Why do the first 100,000 zeroes of the Riemann Zeta function have double-digit sequence count discontinuities at 00,11,22,33,44,55,66,77,88,99?

I was investigating Benford Law type behaviors and was running some analyses on the web page “Andrew Odlyzko: Tables of zeros of the Riemann zeta function” … more specifically the first 100,000 zeroes. The web page is First 100k zeroes of Riemann Zeta function and I simply performed a web browser search for each of the 100 two-digit sequences from 00 to 99.

What I found was as follows:

I understand why the “01” through “09” is in general lower, since leading zeroes do not appear. I also understand the discontinuity from 74 to 75 since the first 100,000 zeros list only goes up to 74921.

But why are multiples of eleven incrementally less frequent than their nearby digit sequences.

Noticed that

C(“01”) is greater than C(“00”),

C(“10”) and C(“12”) are greater than C(“11”),

C(“21”) and C(“23”) are greater than C(“22”), et cetera,

continuing on to

C(“98”) is greater than C(“99”).Seems curious to me.

Any insight here would be appreciated.

**Answer**

This has nothing to do with the Riemann zeta function, but is rather a property of random sequences of digits. (As a rule there’s no reason to assume that there’s anything significant about the decimal expansion of any number, unless there’s some concrete reason to believe otherwise.)

In particular, assume that we have a random sequence of digits where all of 0, 1, 2, …, 9 are equally likely. Then you expect any particular two-digit sequence to occur 1 time in 100 as a pair of consecutive digits. Most of the numbers at the web site you’ve linked to are five digits, then a decimal point, then nine more digits. So in each of these numbers there are 12 possible “slots” where, say, “42” can occur, four before the decimal point and eight after. Across all the numbers there are 1.2 million such “slots” (actually a few less because not all the numbers have five digits before the decimal point) and so you expect (1.2 million)/(100) = 12000 occurrences of “42”. This is what you see in your plot, except for the deviation because the sequence of zeros cuts off at 74931.

However, how should your browser behave if the sequence “111” appears? You’ll find that it counts that as a single occurrence of “11”. But the analysis I just gave would count it twice. So the browser-based counts for a sequence of digits that can overlap itself, such as 11, will be lower. In the literature of combinatorics on words this phenomenon is called “autocorrelation”.

**Attribution***Source : Link , Question Author : phdmba7of12 , Answer Author : Michael Lugo*